Stochastic Programming - Index of

RECOURSE PROBLEMS 187
means of the following problem, where we calculate what is left for the next
random variable:
α r+1
i
= α r i − min{yr+ i
,y r−
i
, 0}. (4.6)
What we are doing here is to find, for each variable, how much ˜ξ r ,intheworst
case, uses of the bound on variable i in the negative direction. That is then
subtracted off what we had before. There are three possibilities. Both (4.4) and
(4.5) may yield non-negative values for the variable y i . In that case nothing
is used of the available “negative bound” α r i .Thenαr+1 i = α r i . Alternatively,
if (4.4) has yi r+ < 0, then it will in the worst case use yi
r+ of the available
“negative bound”. Finally, if (4.5) has y r−
i < 0thenintheworstcaseweuse
y r−
i of the bound. Therefore α r+1
i is what is left for the next random variable.
Similarly, we find
β r+1
i
= βi r − max{yr+ i
,y r−
i
, 0}, (4.7)
where β r+1
i shows how much is still available of bound i in the forward
(positive) direction.
We next increase the counter r by one and repeat (4.4)–(4.7). This takes
care of the piecewise linear functions in ξ.
Note that it is possible to solve (4.4) and (4.5) by parametric linear
programming, thereby getting not just one linear piece above E ˜ξ and one
below, but rather piecewise linearity on both sides. Then (4.6) and (4.7) must
be updated to “worst case” analysis of bound usage. That is simple to do.
Let us turn to our example (4.1). Since we have developed the piecewise
linear upper bound for equality constraints, we shall repeat the problem with
slack variables added explicitly.
φ(ξ 1 ,ξ 2 )=min{2x raw1 +3x raw2 }
s.t. x raw1 + x raw2 + s 1 = 100,
2x raw1 +6x raw2 − s 2 = 180 + ξ 1 ,
3x raw1 +3x raw2 − s 3 = 162 + ξ 2 ,
x raw1 ≥ 0,
x raw2 ≥ 0,
s 1 ≥ 0,
s 2 ≥ 0,
s 3 ≥ 0.
In this setting, what we need to develop is the following:
{ {
d
+
U(ξ 1 ,ξ 2 )=φ(0, 0) + 1 ξ 1 if ξ 1 ≥ 0, d
d − 1 ξ 1 if ξ 1 < 0, + +
2 ξ 2 if ξ 2 ≥ 0,
d − 2 ξ 2 if ξ 2 < 0.
First, we have already calculated φ(0, 0) = 126 with x raw1 = 36, x raw2 =
18 and s 1 = 46. Next, let us try to find d ± 1 . To do that, we need α1 ,